THERMODYNAMICS: Calculate the value of AH (in kJ) for the reaction, 2AR3 (g) + Q (s)→ A2R4 (g) + QR2 (s) given the following hypothetical thermochemical equations: 2AR3 (g) + 2AX (g) → A2R4 (g) + 2XAR (g) AH = -87.2 kJ AX (g) + %R2 (g) XAR (g) AH = -157.3 kJ Q (s) + R2 (g) → QR2 (s) AH = -510.2 kJ (Round off the final answer to ONE decimal place. Do not include the unit.)

THERMODYNAMICS: Calculate the value of AH (in kJ) for the reaction, 2AR3 (g) + Q (s)→ A2R4 (g) + QR2 (s) given the following hypothetical thermochemical equations: 2AR3 (g) + 2AX (g) → A2R4 (g) + 2XAR (g) AH = -87.2 kJ AX (g) + %R2 (g) XAR (g) AH = -157.3 kJ Q (s) + R2 (g) → QR2 (s) AH = -510.2 kJ (Round off the final answer to ONE decimal place. Do not include the unit.)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) – 4Q2 (g) + 3J2X (1) AH = -1,070 kJ 4QJ3 (g) + 3X2 (g) → 202 (g) + 6J2X (1) AH = -1,580.2 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + % X2 (g) –→ Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit.)
2. จงหาค่า AH ของปฏิกิริยานี้ จากข้อมูลที่กำหนดให้ 2C (s) + 2 H,0 (g) → CH, (g) + CO,(g) Determine the standard enthalpy change for this reaction from the following standard enthalpies of reaction: (1) C(s) + H,O (g) → CÓ (g) + H, (g) AH° = 131.3kJ (2) CO (g) + H2O (g) → CO2 (g) + H2 (g) ΔΗΟ = -41.2 kJ (3) CH, (g) + H,O (g) → 3H, (g) + CO (g) AH° = 206.1kJ ชื่อ-สกุล .วันที่ เลขที่ เรหัส.. (0202104) จงหาค่า AH และ As ของปฏิกิริยานี้ จากข้อมูลที่กำหนดให้ 3 ปฏิกิริยาการสังเคราะห์ด้วยแสงเกิดขึ้นดังสมการ 6 CO, (g) + 6 H,0 (1) CH2O6 (S) + 6 0, (g) กำหนดค่าที่สภาวะมาตรฐานของ AH และ S ของสารแต่ละตัว ดังนี้ AH°, (kJ mol-') s° (JK-1 mol-') สาร | CH,206 (s) -1,273 212 O, (g) 205 2/2 CO, (g) -393 214 H,0 (1) -285 70
4. Hess's Law Calculate AH for the reaction: 2 C(s) + H2(g) → C:H2(g), given the following chemical equations and their respective enthalpy changes: C2H2(g)+02(g)→2CO2(g)+H20(1) AH, = -1299.6 kJ (1) %3D C(s)+02(g)→CO2(g) AH, = -393.5 kJ (2) H2(g)+502(g)→H20(1) AH3 = -285.8 kJ (3)
Determine the enthalpy of the reaction: 3Fe,O; (s) + CO (g) → CO, (g) + 2Fe;O4 (s), given the following reactions. 38. Fe,O3 (s) + 3CO (g) → 2Fe (s) + 3CO, (g) AH= -28.0 kJ 3Fe (s) + 4CO2 (s) → 4CO (g) + Fe;O4 (s) AH =+12.5 kJ A) 40.5 +109 -59.0 -109
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) – 4Q2 (9) + 3J2X (1) AH = -1,017.3 kJ 4QJ3 (g) + 3X2 (g) → 202 (9) + 6J2X (1) AH = -1,548.8 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + % X2 (g) –→ Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit.)
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (l) ΔH = -1,004 kJ 4QJ3 (g) + 3X2 (g) → 2Q2 (g) + 6J2X (l) ΔH = -1,520.1 kJ Calculate the value of ΔH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) → Q2X (g)
4) Given the data N2(8) + O2(8) 2 NO(8) AH = +180.7 kJ 2 NO(8) + O2(8) · 2 NO,(8) 2 N,0(8) – 2 N2(8) + O2(8) AH = -163.2 kJ AH = -113.1 kJ | use Hess's law to calculate AH for the reaction N20(8) + NO2(8) –→ 3 NO(8)
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) – 4Q2 (g) + 3J2X (1) AH = -1,096.4 kJ 4QJ3 (g) + 3X2 (g) → 202 (g) + 6J2X (1) AH = -1,548.8 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + % X2 (g) –→ Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit,.)
THERMODYNAMICS: Calculate the value of AH (in kJ) for the reaction, 2AR3 (9) + Q (s) → A2R4 (g) + QR2 (s) given the following hypothetical thermochemical equations: 2AR3 (g) + 2AX (g) → A2R4 (9) + 2XAR (g) ΔΗ = -88.4 kJ AX (g) + ½R2 (g) → XAR (g) AH = -164.8 kJ Q (s) + R2 (g) –→ QR2 (s) AH = -520.4 kJ (Round off the final answer to ONE decimal place. Do not include the unit.)
Study the following thermochemical data very carefully: Clelg) lag) ICI(g) Ia(s) 2C(g) 21(g) I(g) + CI(g) la(g) AH = 242.3 kJ AH = 151.0 kJ AH = 211.3 kJ AH = 62.8 kJ I2(s) reacts with Clag) to produce ICI(g). Write the chemical equation to produce ONLY one mole of ICI(g)
to calculate the true enthalpies of these reactions by the Hess’ law, as: ΔH°(rxn) = Σ[ν × ΔHf°(product)] – Σ[ν × ΔHf°(reactant)] for: Mg(s) + ½O2(g) → MgO(s)
（1）Consider the reaction: 2A (g) + 3 B (g) → 2 C (g) ΔHrxn = +254.3 kJ What will be the enthalpy change (in kJ) if 0.471 mol B reacts in excess A? （2）Consider the reaction: C (s) + O2 (g) → CO2 (g) ΔHrxn = -393.5 kJ What mass of carbon (in g) must be reacted via this mechanism to release 581.2 kJ of heat?